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Straight in LC oscillation, the sum of energies stored in capacitor & theinductor is constant in time. |
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Answer» Solution :i. During LC oscilations in LC circuits, the ENERGY of the system oscillates between the electric field of the capacitor and the magnetic field of the INDUCTOR. ii. Althogh, these two forms of energy vary with time, the total energy REMAINS constant. It means that with the law of constant. It accordance with the law of conservation of energy. Total energy, `U=U_(E)+U_(B)=(q^(2))/(2C)+(1)/(2)Li^(2)` iii. consider 3 different stages of LC oscillations and calculate the total energy of the system. Case (i)When the charge in the capacitor, q `Q_(m)=` and the current throughthe inductor, i = 0, the total energy is GIVEN by `U=(Q_(m)^(2))/(2C)+0=(Q_(m)^(2))/(2C)` The total energy is wholly electrical. Case (ii)When charge = 0 , current `I_(m),` `U=0+(1)/(2)LI_(m)^(2)=(1)/(2)LI_(m)^(2)` `=(L)/(2)xx((Q_(m)^(2))/(LC))"since Im"=Q_(m)omega=(Q_(m))/(sqrt(LC))` `=(Q_(m)^(2))/(2C)` Case (iii)When charge = q, current = i, the total energy is `U=(q^(2))/(2C)+(1)/(2)Li^(2)` IV. Since `q=Q_(m)=cosomegat,i=(dq)/(dt)=Q_(m)=omegasinomegat.` The negative sign in current indicates that the charge in the capacitor decreases with time. `U=(Q_(m)^(2)cos^(2)omegat)/(2C)+(Lomega^(2)Q_(m)^(2)sin^(2)omegat)/(2)` `U=(Q_(m)^(2)cos^(2)omegat)/(2C)+(Lomega_(m)^(2)Q_(m)^(2)sin^(2)omegat)/(2)" since "omega^(2)=(1)/(LC)` `=(Q_(m)^(2))/(2C)(cos^(2)omegat+sin^(2)omegat)` `U=(Q_(m)^(2))/(2C)` Frome above three cases, it is clear that the total energy of the system remains constant. |
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